Problem: Ikta has an extraordinary feeling for undirected graphs. Ikta likes the undirected graph G and its two-point s, t pair (G, s, t) that has the greatest "beauty". The "beauty" of a pair (G, s, t) is the edge e = \\ {u, v \\} (u and v are two different points of G), and the shortest path from s to t in G. Is the number of things whose length is greater than the length of the shortest path from s to t in the undirected graph with g plus e.

Your job is to write a program that seeks its "beauty" given a pair (G, s, t).



Input

The input is given in the following format.

> N M s t
> x1 y1
> ...
> xi yi
> ...
> xM yM
>

First, the number of vertices, the number of edges, and the integers N, M, s, t representing the two vertices of the undirected graph are input. From the 2nd line to the M + 1th line, two vertices connected by edges are input. (However, let the set of G vertices be \\ {1, ..., N \\}.)

Constraints

Each variable being input satisfies the following constraints.

* 2 ≤ N ≤ 100,000
* 1 ≤ M ≤ 300,000
* 1 ≤ s, t, xi, yi ≤ N
* Different from s and t
* Guaranteed to reach t from s

Output

When the given graph is G, output the "beauty" of the set (G, s, t) in one line.

Examples

Input

3 2 1 3
1 2
2 3


Output

1


Input

9 8 7 8
2 6
4 9
8 6
9 2
3 8
1 8
8 5
7 9


Output

7


Input

4 3 1 4
1 2
3 4
4 1


Output

0


Input

9 7 8 9
9 6
6 5
3 6
3 7
2 5
8 5
1 4


Output

2

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
#define SZ(x) (int)(x.size())
#define REP(i, n) for(int i=0;i<(n);++i)
#define FOR(i, a, b) for(int i=(a);i<(b);++i)
#define RREP(i, n) for(int i=(int)(n)-1;i>=0;--i)
#define RFOR(i, a, b) for(int i=(int)(b)-1;i>=(a);--i)
#define ALL(a) a.begin(),a.end()
#define DUMP(x) cerr<<#x<<" = "<<(x)<<endl
#define DEBUG(x) cerr<<#x<<" = "<<(x)<<" (L"<<__LINE__<<")"<< endl;
using ll = long long;
using vi = vector<int>;
using vvi = vector<vi>;
using vll = vector<ll>;
using vvll = vector<vll>;
using P = pair<int, int>;
const double eps = 1e-8;
const ll MOD = 1000000007;
const int INF = INT_MAX / 2;
const ll LINF = LLONG_MAX / 2;
template <typename T1, typename T2>
bool chmax(T1 &a, const T2 &b) {
    if (a < b) { a = b; return true; }
    return false;
}
template <typename T1, typename T2>
bool chmin(T1 &a, const T2 &b) {
    if (a > b) { a = b; return true; }
    return false;
}
template<typename T1, typename T2>
ostream &operator<<(ostream &os, const pair<T1, T2> &p) {
    os << p.first << ":" << p.second;
    return os;
}
template<typename T1, typename T2>
ostream &operator<<(ostream &os, const map<T1, T2> &mp) {
    os << "{";
    int a = 0;
    for (auto &tp : mp) {
        if (a) os << ", "; a = 1;
        os << tp;
    }
    return os << "}";
}
template<typename T>
ostream &operator<<(ostream &os, const vector<T> &v) {
    os << "[";
    REP(i, SZ(v)) {
        if (i) os << ", ";
        os << v[i];
    }
    return os << "]";
}


int main() {
    cin.tie(0);
    ios::sync_with_stdio(false);
    cout << fixed << setprecision(10);

    int N, M, s, t;
    cin >> N >> M >> s >> t;
    --s, --t;
    vvi graph(N);
    REP(i, M) {
        int u, v; cin >> u >> v; --u, --v;
        graph[u].push_back(v);
        graph[v].push_back(u);
    }

    vi distS(N, INF), distT(N, INF);
    {
        queue<int> que;
        que.push(s);
        distS[s] = 0;
        while (!que.empty()) {
            int v = que.front(); que.pop();
            for (int u : graph[v]) {
                if (chmin(distS[u], distS[v] + 1)) {
                    que.push(u);
                }
            }
        }
    }
    {
        queue<int> que;
        que.push(t);
        distT[t] = 0;
        while (!que.empty()) {
            int v = que.front(); que.pop();
            for (int u : graph[v]) {
                if (chmin(distT[u], distT[v] + 1)) {
                    que.push(u);
                }
            }
        }
    }

    map<int,ll> cntS, cntT;
    for (int d : distS) ++cntS[d];
    for (int d : distT) ++cntT[d];

    int D = distS[t];
    ll ans = 0;
    for (auto &tp : cntS) {
        ans += tp.second * cntT[D - 2 - tp.first];
    }
    cout << ans << endl;

    return 0;
}